Question 221:
moderateAn alternating electric field, of frequency $\nu$, is applied across the dees (radius $= R$) of a cyclotron that is being used to accelerate protons ($\text{mass} = m$). The operating magnetic field ($B$) used in the cyclotron and the kinetic energy ($K$) of the proton beam, produced by it, are given by:
(2012 Pre)
Cyclotron frequency is $\nu = \frac{eB}{2\pi m}$, giving $B = \frac{2\pi m\nu}{e}$. Maximum kinetic energy is $K = \frac{e^2B^2R^2}{2m} = 2m\pi^2\nu^2R^2$.
Question 222:
easyA uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron:
(2011 Pre)
The magnetic force is zero because velocity and magnetic field are parallel. The electric force acts opposite to the electron's motion since it is negatively charged, causing its speed to decrease.
Question 223:
easyA square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $vec{F}$, the net force on the remaining three arms of the loop is:
(2010 Pre)
The net magnetic force on a closed current loop in a uniform magnetic field is always zero.
Therefore, $\vec{F} + \vec{F}_{\text{remaining}} = 0$, which gives $\vec{F}_{\text{remaining}} = -\vec{F}$.
Question 224:
moderateA beam of cathode rays is subjected to crossed Electric (E) and Magnetic field (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by (where V is the potential difference between cathode and anode):
(2010 Pre)
For undeflected motion, velocity is $v = \frac{E}{B}$.
Equating kinetic energy to electrical work gives $\frac{1}{2}mv^2 = eV$.
Substituting velocity yields specific charge $\frac{e}{m} = \frac{E^2}{2VB^2}$.
Question 225:
moderateA charge ‘q’ moves in a region where electric field and magnetic field both exist, then force on it is:
(2002)
The total electromagnetic force on a moving charge in both electric and magnetic fields is the Lorentz force.
It is the vector sum of the electric force $q\vec{E}$ and the magnetic force $q(\vec{V} \times \vec{B})$.
Question 226:
easyTwo long parallel wires are at a distance of $1\text{ m}$. If both of them carry one ampere of current in same direction, then the force of attraction on unit length of the wires will be:
(1998)
The force per unit length between two parallel current-carrying wires is $\frac{F}{l} = \frac{mu_0 I_1 I_2}{2\pi d}$.
Substituting $I_1 = 1\text{ A}$, $I_2 = 1\text{ A}$, and $d = 1\text{ m}$ results in $2 \times 10^{-7}\text{ N/m}$.
Question 227:
easyA beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength $20\text{ Vm}^{-1}$ and $0.5\text{ T}$ at right angles to the direction of motion of the electrons. What is the velocity of the electrons?
(1996)
For undeflected particle motion in perpendicular fields, magnetic force balances electric force ($qE = qvB$).
Thus, velocity $v = \frac{E}{B} = \frac{20}{0.5} = 40\text{ ms}^{-1}$.
Question 228:
moderateA straight wire of length $0.5\text{ metre}$ and carrying a current of $1.2\text{ ampere}$ is placed in uniform magnetic field of induction $2\text{ tesla}$. The magnetic field is perpendicular to the length of the wire. The force on the wire is:
(1992)
The magnetic force on a current-carrying wire is given by $F = I l B sintheta$.
Substituting $I = 1.2text{ A}$, $l = 0.5text{ m}$, $B = 2text{ T}$, and $theta = 90^circ$ yields $F = 1.2text{ N}$.
Question 229:
moderateA coil in the shape of an equilateral triangle of side $L$ is suspended between the pole pieces of a permanent magnet such that $B$ is in plane of the coil. If due to a current $i$ in the triangle a torque $tau$ acts on it, the side $L$ of the triangle is :
(2005)
The magnetic moment is $M = i A = i \left( \frac{\sqrt{3}}{4} L^2 \right)$. The torque is $\tau = M B = i \frac{sqrt{3}}{4} L^2 B$. Solving for $L$ yields $2 \left[ \frac{\tau}{\sqrt{3}Bi} \right]^{1/2}$.
Question 230:
easyA current loop in a magnetic field :
(2013)
A current loop in a uniform magnetic field experiences zero net force and can be in equilibrium when its magnetic moment is parallel (stable equilibrium) or antiparallel (unstable equilibrium) to the field.